Not exacly the place for it... Math help

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anthony88guy
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Not exacly the place for it... Math help

Post by anthony88guy »

Instead of finding a math forum, registering and that whole tid bit, I figured ill try my query here first. Even if someone could just tell me whats this called besides transversal lines, I would have some luck finding something helpful on the web.

So here we go a geometry question:
Image
(not to scale)

As stated in the image AC and DB are parallel lines.

AB = 14
AC = 3
DC = 4


Now we have to find AE.

I remember having to set up a proportion? Something along the lines of:
Image

But we don't have enough information to do so.

Any help would be greatly appreciated.

Thanks.
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feyd
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Post by feyd »

Looks like an application of the Law of Sines and Law of Cosines.
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Oren
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Post by Oren »

ACE =~ BDE, thus:

AC/DB = AE/BE

AC = 3
DB = 4

=>

3/4 = AE/BE

Let's write AE = x
AE = x; AB = 14 => BE = AB - AE = 14 - x

=>

3/4 = x/(14 - x)

=>

AE = x = 6

:wink:
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superdezign
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Post by superdezign »

Oren wrote:AE = x; AB = 14 => BE = AB - AE = 14 - x
^_^ I need to get back into math class. That thought didn't even come to mind.
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Oren
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Post by Oren »

Actually it should be like this (but who cares):

1. Denote: AE = x
2. AB = 14
3. BE = AB - AE

=>

BE = 14 - x

P.S I didn't give full proof as this is not a math test (e.g I didn't prove that ACE =~ BDE since it's too easy).
anthony88guy
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Post by anthony88guy »

Thank you Oren.

I knew it was a proportion problem just couldn't figure out how to set it up.
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RobertGonzalez
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Post by RobertGonzalez »

This also fits into the category of similarity. The fact that AC||DB means that triangle ACE is similar to triangle BDE (provided segment AB and segment CD are linear). Given the similarity, you can set up proportionate equality to solve, as Oren did.
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Oren
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Post by Oren »

Everah wrote:This also fits into the category of similarity. The fact that AC||DB means that triangle ACE is similar to triangle BDE (provided segment AB and segment CD are linear). Given the similarity, you can set up proportionate equality to solve, as Oren did.
Everah, that's exactly what I used. "ACE =~ BDE" means "triangle ACE is similar to triangle BDE" :wink:

What do you mean by the way when you say "provided segment AB and segment CD are linear"? AC||DB that's enough.
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RobertGonzalez
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Post by RobertGonzalez »

I know that is what you meant. I just wasn't sure the OP knew that =~ means similar.

And I suppose by linear I was just complicating things. I was thinking that if the intersecting segments were curves, then it would throw similarity out the window, but I think I am just being weird (and showing my age a little. I haven't been deeply involved in Geometry for about 18 years now).
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Oren
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Post by Oren »

Everah wrote:I haven't been deeply involved in Geometry for about 18 years now.
Wow, and yet you remembered? That's great man... seriously. I can be a master in many different things, but if I don't use them for few months then I totally forget them. I'm really amazed that you remembered that after nearly 18 years :bow:
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RobertGonzalez
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Post by RobertGonzalez »

Math was the one thing that I really enjoyed throughout my education. I took math all the way through linear equations and differential calculus. Of course, that was 13 years ago (when I completed my math learning). Geometry was my sophomore year in high school.

PS I just realized how old that makes me sound. Image
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patrikG
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Post by patrikG »

Are you guys helping some guy with his homework again? ;)
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RobertGonzalez
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Post by RobertGonzalez »

Hell no. I purposely muddled things up to make him go back to his text book and learn it the right way. ;)
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